Chapter 3 Playing with Numbers - Mentorminutesmentorminutes.com/Class VI Mathematics Chapter...

Class VI

Chapter 3 – Playing with Numbers

Maths

Exercise 3.1

Question 1:

Write all the factors of the following numbers:

(a) 24 (b) 15 (c) 21

(d) 27 (e) 12 (f) 20

(g) 18 (h) 23 (i) 36

Answer:

(a) 24

24 = 1 × 24 24 = 2 × 12 24 = 3 × 8

24 = 4 × 6 24 = 6 × 4

∴Factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24

(b) 15

15 = 1 × 15 15 = 3 × 5 15 = 5 × 3

∴Factors of 15 are 1, 3, 5, and 15

(c) 21

21 = 1 × 21 21 = 3 × 7 21 = 7 × 3


(d) 27

27 = 1 × 27 27 = 3 × 9 27 = 9 × 3


Page 1 of 41

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Class VI


Maths

(e) 12

12 = 1 × 12 12 = 2 × 6 12 = 3 × 4 12 = 4 × 3

∴Factors of 12 are 1, 2, 3, 4, 6, and 12

(f) 20

20 = 1 × 20 20 = 2 × 10 20 = 4 × 5 20 = 5 × 4


(g) 18

18 = 1 × 18 18 = 2 × 9 18 = 3 × 6 18 = 6 × 3


(h) 23

23 = 1 × 23 23 = 23 × 1

∴ Factors of 23 are 1 and 23

(i) 36

36 = 1 × 36 36 = 2 × 18 36 = 3 × 12 36 = 4 × 9

36 = 6 × 6

∴Factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36

Page 2 of 41


Column 1 Column 2

(i) 35 (a) Multiple of 8

(ii) 15 (b) Multiple of 7

(iii) 16 (c) Multiple of 70

(iv) 20 (d) Factor of 30

(v) 25 (e) Factor of 50

Class VI


Maths

Question 2:

Write first five multiplies of:

(a) 5 (b) 8 (c) 9

Answer:

(a) 5 × 1 = 5 5 × 2 = 10 5 × 3 = 15 5 × 4 = 20 5 × 5 = 25

∴ The required multiples are 5, 10, 15, 20, and 25.

(b) 8 × 1 = 8 8 × 2 = 16 8 × 3 = 24 8 × 4 = 32 8 × 5 = 40


(c) 9 × 1 = 9 9 × 2 = 18 9 × 3 = 27 9 × 4 = 36 9 × 5 = 45


Question 3:

Match the items in column 1 with the items in column 2.

Page 3 of 41


Column 1 Column 2

(i) 35 (b) Multiple of 7

(ii) 15 (d) Factor of 30

(iii) 16 (a) Multiple of 8

(iv) 20 (f) Factor of 20

(v) 25 (e) Factor of 50

Class VI

-

Chapter 3 – Playing with Numbers (f) Factor of 20

Maths

Answer:

Question 4:

Find all the multiples of 9 up to 100.

Answer:

9 × 1 = 9 9 × 2 = 18 9 × 3 = 27 9 × 4 = 36 9 × 5 = 45

9 × 6 = 54 9 × 7 = 63 9 × 8 = 72 9 × 9 = 81 9 × 10 = 90

9 × 11 = 99

Therefore, the multiples of 9 up to 100 are

9, 18, 27, 36, 45, 54, 63, 72, 81, 90, and 99

Page 4 of 41


Class VI


Maths

Exercise 3.2

Question 1:

What is the sum of any two (a) Odd numbers? (b) Even numbers?

Answer:

(a) The sum of two odd numbers is even.

e.g., 1 + 3 = 4

13 + 19 = 32

(b) The sum of two even numbers is even.

e.g., 2 + 4 = 6

10 + 18 = 28

Question 2:

State whether the following statements are True or False:

(a) The sum of three odd numbers is even.

(b) The sum of two odd numbers and one even number is even.

(c) The product of three odd numbers is odd.

(d) If an even number is divided by 2, the quotient is always odd.

(e) All prime numbers are odd.

(f) Prime numbers do not have any factors.

(g) Sum of two prime numbers is always even.

(h) 2 is the only even prime number.

(i) All even numbers are composite numbers.

(j) The product of two even numbers is always even.

Answer:

(a) False 3 + 5 + 7 = 15, i.e., odd

(b) True 3 + 5 + 6 = 14, i.e., even

(c) True 3 × 5 × 7 = 105, i.e., odd

(d) False 4 ÷ 2 = 2, i.e., even

(e) False 2 is a prime number and it is also even

(f) False 1 and the number itself are factors of the number

(g) False 2 + 3 = 5, i.e., odd

(h) True

(i) False 2 is a prime number

(j) True 2 × 4 = 8, i.e., even

Page 5 of 41

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Class VI


Maths

Question 3:

The numbers 13 and 31 are prime numbers. Both these numbers have same digits 1 and

3. Find such pairs of prime numbers up to 100.

Answer:

17, 71

37, 73

79, 97

Question 4:

Write down separately the prime and composite numbers less than 20.

Answer:

Prime numbers less than 20 are

2, 3, 5, 7, 11, 13, 17, 19

Composite numbers less than 20 are

4, 6, 8, 9, 10, 12, 14, 15, 16, 18

Question 5:

What is the greatest prime number between 1 and 10?

Answer:

Prime numbers between 1 and 10 are 2, 3, 5, and 7. Among these numbers, 7 is the

greatest.

Question 6:

Express the following as the sum of two odd primes.

(a) 44 (b) 36 (c) 24 (d) 18

Answer:

(a) 44 = 37 + 7

(b) 36 = 31 + 5

(c) 24 = 19 + 5

(d) 18 = 11 + 7

Question 7:

Give three pairs of prime numbers whose difference is 2.

[Remark: Two prime numbers whose difference is 2 are called twin primes].

Answer:

3, 5

41, 43

Page 6 of 41



Class VI


Maths

71, 73

Question 8:

Which of the following numbers are prime?

(a) 23 (b) 51 (c) 37 (d) 26

Answer:

(a) 23 23 = 1 × 23 23 = 23 × 1

23 has only two factors, 1 and 23. Therefore, it is a prime number.

(b) 51 51 = 1 × 51 51 = 3 × 17

51 has four factors, 1, 3, 17, 51. Therefore, it is not a prime number. It is a composite

number.

(c) 37

It has only two factors, 1 and 37. Therefore, it is a prime number.

(d) 26

26 has four factors (1, 2, 13, 26). Therefore, it is not a prime number. It is a composite

number.

Question 9:

Write seven consecutive composite numbers less than 100 so that there is no prime

number between them.

Answer:

Between 89 and 97, both of which are prime numbers, there are 7 composite numbers.

They are

90, 91, 92, 93, 94, 95, 96

Numbers Factors

90 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90

91 1, 7, 13, 91

92 1, 2, 4, 23, 46, 92

93 1, 3, 31, 93

94 1, 2, 47, 94

95 1, 5, 19, 95

96 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96

Page 7 of 41


Class VI


Maths

Question 10:

Express each of the following numbers as the sum of three odd primes:

(a) 21 (b) 31 (c) 53 (d) 61

Answer:

(a) 21 = 3 + 7 + 11

(b) 31 = 5 + 7 + 19

(c) 53 = 3 + 19 + 31

(d) 61 = 11 + 19 + 31

Question 11:

Write five pairs of prime numbers less than 20 whose sum is divisible by 5.

(Hint: 3 + 7 = 10)

Answer:

2+3=5

2 + 13 = 15

3 + 17 = 20

7 + 13 = 20

19 + 11 = 30

Question 12:

Fill in the blanks:

(a) A number which has only two factors is called a _______.

(b) A number which has more than two factors is called a _______.

(c) 1 is neither _______ nor _______.

(d) The smallest prime number is _______.

(e) The smallest composite number is _______.

(f) The smallest even number is _______.

Answer:

(a) Prime number

(b) Composite number

(c) Prime number, composite number

(d) 2

(e) 4

(f) 2

Page 8 of 41


Class VI


Maths

Exercise 3.3

Question 1:

Using divisibility tests, determine which of the following numbers are divisible by 2; by

3; by 4; by 5; by 6; by 8; by 9; by 10; by 11 (say, yes or no):

Divisible by

Number

2 3 4 5 6 8 9 10 11

128

990

1586

275

6686

639210

429714

2856

3060

406839

Yes

…

…

…

…

…

…

…

…

…

No

…

…

…

…

…

…

…

…

…

Yes

…

…

…

…

…

…

…

…

…

No

…

…

…

…

…

…

…

…

…

No

…

…

…

…

…

…

…

…

…

Yes

…

…

…

…

…

…

…

…

…

No

…

…

…

…

…

…

…

…

…

No

…

…

…

…

…

…

…

…

…

No

…

…

…

…

…

…

…

…

…

Answer:

Numbers

990

1586

275

2

Yes

Yes

No

3

Yes

No

No

4

No

No

No

5

Yes

No

Yes

6

Yes

No

No

8

No

No

No

9

Yes

No

No

10

Yes

No

No

11

Yes

No

Yes

Page 9 of 41


6686 Yes No No No No No No No No

639210 Yes Yes No Yes Yes No No Yes Yes

429714 Yes Yes No No Yes No Yes No No

2856 Yes Yes Yes No Yes Yes No No No

3060 Yes Yes Yes Yes Yes No Yes Yes No

406839 No Yes No No No No No No No

Class VI


Maths

Question 2:

Using divisibility tests, determine which of the following numbers are divisible by 4; by

8:

(a) 572 (b) 726352 (c) 5500 (d) 6000

(e) 12159 (f) 14560 (g) 21084 (h) 31795072

(i) 1700 (j) 2150

Answer:

(a) 572

The last two digits are 72. Since 72 is divisible by 4, the given number

is also divisible by 4.

The last three digits are 572. Since 572 is not divisible by 8, the given number is also

not divisible by 8.

(b) 726352

The last two digits are 52. As 52 is divisible by 4, the given number is also divisible by 4.

The last three digits are 352. Since 352 is divisible by 8, the given number is also

divisible by 8.

(c) 5500

Since last two digits are 00, it is divisible by 4.

The last 3 digits are 500. Since 500 is not divisible by 8, the given number is also not

divisible by 8.

(d) 6000

Since the last 2 digits are 00, the given number is divisible by 4.

Page 10 of 41



Class VI


Maths

Since the last 3 digits are 000, the given number is divisible by 8.

(e) 12159

The last 2 digits are 59. Since 59 is not divisible by 4, the given number is also not

divisible by 4.

The last 3 digits are 159. Since 159 is not divisible by 8, the given number is not

divisible by 8.

(f) 14560

The last two digits are 60. Since 60 is divisible by 4, the given number is divisible by 4.

The last 3 digits are 560. Since 560 is divisible by 8, the given number is divisible by 8.

(g) 21084


The last three digits are 084. Since 084 is not divisible by 8, the given number is not

divisible by 8.

(h) 31795072


The last three digits are 072. Since 072 is divisible by 8, the given number is divisible by

8.

(i) 1700



divisible by 8.

(j) 2150

The last two digits are 50. Since 50 is not divisible by 4, the given number is not

divisible by 4.


divisible by 8.

Question 3:

Using divisibility tests, determine which of following numbers are divisible by 6:

(a) 297144 (b) 1258 (c) 4335 (d) 61233

(e) 901352 (f) 438750 (g) 1790184 (h) 12583

(i) 639210 (j) 17852

Answer:

(a) 297144

Page 11 of 41



Class VI


Maths

Since the last digit of the number is 4, it is divisible by 2.

On adding all the digits of the number, the sum obtained is 27. Since 27 is divisible by 3,

the given number is also divisible by 3.

As the number is divisible by both 2 and 3, it is divisible by 6.

(b) 1258


On adding all the digits of the number, the sum obtained is 16. Since 16 is not divisible

by 3, the given number is also not divisible by 3.

As the number is not divisible by both 2 and 3, it is not divisible by 6.

(c) 4335

The last digit of the number is 5, which is not divisible by 2. Therefore, the given number

is also not divisible by 2.




(d) 61233

The last digit of the number is 3, which is not divisible by 2. Therefore, the given number

is also not divisible by 2.




(e) 901352





(f) 438750





(g) 1790184


Page 12 of 41



Class VI


Maths




(h) 12583

Since the last digit of the number is 3, it is not divisible by 2.




(i) 639210





(j) 17852





Question 4:

Using divisibility tests, determine which of the following numbers are divisible by 11:

(a) 5445 (b) 10824 (c) 7138965 (d) 70169308

(e) 10000001 (f) 901153

Answer:

(a) 5445

Sum of the digits at odd places = 5 + 4 = 9

Sum of the digits at even places = 4 + 5 = 9

Difference = 9 − 9 = 0

As the difference between the sum of the digits at odd places and the sum of the digits

at even places is 0, therefore, 5445 is divisible by 11.

(b) 10824

Sum of the digits at odd places = 4 + 8 + 1 = 13

Sum of the digits at even places = 2 + 0 = 2


Page 13 of 41



Class VI


Maths

The difference between the sum of the digits at odd places and the sum of the digits at

even places is 11, which is divisible by 11. Therefore, 10824 is divisible by 11.

(c) 7138965

Sum of the digits at odd places = 5 + 9 + 3 + 7 = 24

Sum of the digits at even places = 6 + 8 + 1 = 15


The difference between the sum of the digits at odd places and the sum of digits at even

places is 9, which is not divisible by 11. Therefore, 7138965 is not divisible by 11.

(d) 70169308

Sum of the digits at odd places = 8 + 3 + 6 + 0 = 17

Sum of the digits at even places = 0 + 9 + 1 + 7 = 17




(e) 10000001

Sum of the digits at odd places = 1

Sum of the digits at even places = 1




(f) 901153




The difference between the sum of the digits at odd places and the sum of the digits at

even places is 11, which is divisible by 11. Therefore, 901153 is divisible by 11.

Question 5:

Write the smallest digit and the greatest digit in the blank space of each of the following

numbers so that the number formed is divisible by 3:

(a) ___6724 (b) 4765 ___2

Answer:

(a) _6724

Sum of the remaining digits = 19

Page 14 of 41



Class VI


Maths

To make the number divisible by 3, the sum of its digits should be divisible by 3.

The smallest multiple of 3 which comes after 19 is 21.

Therefore, smallest number = 21 − 19 = 2

Now, 2 + 3 + 3 = 8

However, 2 + 3 + 3 + 3 = 11

If we put 8, then the sum of the digits will be 27 and as 27 is divisible by 3, the number

will also be divisible by 3.

Therefore, the largest number is 8.

(b) 4765_2

Sum of the remaining digits = 24

To make the number divisible by 3, the sum of its digits should be divisible by 3. As 24 is

already divisible by 3, the smallest number that can be placed here is 0.

Now, 0 + 3 = 3

3+3=6

3+3+3=9

However, 3 + 3 + 3 + 3 = 12

If we put 9, then the sum of the digits will be 33 and as 33 is divisible by 3, the number

will also be divisible by 3.

Therefore, the largest number is 9.

Question 6:

Write a digit in the blank space of each of the following numbers so that the number

formed is divisible by 11:

(a) 92 ___ 389 (b) 8 ___9484

Answer:

(a) 92_389

Let a be placed in the blank.


Sum of the digits at even places = 8 + a + 9 = 17 + a

Difference = 17 + a − 14 = 3 + a

For a number to be divisible by 11, this difference should be zero or a multiple of 11.

If 3 + a = 0, then

a =−3

However, it cannot be negative.

Page 15 of 41



Class VI


Maths

A closest multiple of 11, which is near to 3, has to be taken. It is 11itself.

3 + a = 11

a =8

Therefore, the required digit is 8.

(b) 8_9484

Let a be placed in the blank.

Sum of the digits at odd places = 4 + 4 + a = 8 + a


Difference = 25 − (8 + a)

= 17 − a

For a number to be divisible by 11, this difference should be zero or a multiple of 11.

If 17 − a = 0, then

a = 17

This is not possible.

A multiple of 11 has to be taken. Taking 11, we obtain

17 − a = 11

a =6

Therefore, the required digit is 6.

Page 16 of 41


Class VI


Maths

Exercise 3.4

Question 1:

Find the common factors of:

(a) 20 and 28 (b) 15 and 25

(c) 35 and 50 (d) 56 and 120

Answer:

(a) Factors of 20 = 1, 2, 4, 5, 10, 20

Factors of 28 = 1, 2, 4, 7, 14, 28

Common factors = 1, 2, 4

(b) Factors of 15 = 1, 3, 5, 15

Factors of 25 = 1, 5, 25

Common factors = 1, 5

(c) Factors of 35 = 1, 5, 7, 35

Factors of 50 = 1, 2, 5, 10, 25, 50


(d) Factors of 56 = 1, 2, 4, 7, 8, 14, 28, 56

Factors of 120 = 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120

Common factors = 1, 2, 4, 8

Question 2:

Find the common factors of:

(a) 4, 8 and 12 (b) 5, 15 and 25

Answer:

(a) 4, 8, 12

Factors of 4 = 1, 2, 4

Factors of 8 = 1, 2, 4, 8

Factors of 12 = 1, 2, 3, 4, 6, 12

Common factors = 1, 2, 4

(b) 5, 15, and 25

Factors of 5 = 1, 5

Factors of 15 = 1, 3, 5, 15

Factors of 25 = 1, 5, 25


Question 3:

Page 17 of 41



Class VI


Maths

Find first three common multiples of:

(a) 6 and 8 (b) 12 and 18

Answer:

(a) 6 and 8

Multiple of 6 = 6, 12, 18, 24, 30…..

Multiple of 8 = 8, 16, 24, 32……

3 common multiples = 24, 48, 72

(b) 12 and 18

Multiples of 12 = 12, 24, 36, 48

Multiples of 18 = 18, 36, 54, 72

3 common multiples = 36, 72, 108

Question 4:

Write all the numbers less than 100 which are common multiples of 3 and 4.

Answer:

Multiples of 3 = 3, 6, 9, 12, 15…

Multiples of 4 = 4, 8, 12, 16, 20…

Common multiples = 12, 24, 36, 48, 60, 72, 84, 96

Question 5:

Which of the following numbers are co-prime?

(a) 18 and 35 (b) 15 and 37 (c) 30 and 415

(d) 17 and 68 (e) 216 and 215 (f) 81 and 16

Answer:

(a) Factors of 18 = 1, 2, 3, 6, 9, 18

Factors of 35 = 1, 5, 7, 35

Common factor = 1

Therefore, the given two numbers are co-prime.

(b) Factors of 15 = 1, 3, 5, 15

Factors of 37 = 1, 37

Common factors = 1


(c) Factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30

Factors of 415 = 1, 5, 83, 415


Page 18 of 41



Class VI


Maths

As these numbers have a common factor other than 1, the given two numbers are not

co-prime.

(d) Factors of 17 = 1, 17

Factors of 68 = 1, 2, 4, 17, 34, 68


As these numbers have a common factor other than 1, the given two numbers are not

co-prime.

(e) 216 and 215

Factors of 216 = 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 27, 36, 54, 72, 108, 216

Factors of 215 = 1, 5, 43, 215

Common factors = 1


(f) 81 and 16

Factors of 81 = 1, 3, 9, 27, 81

Factors of 16 = 1, 2, 4, 8, 16

Common factors = 1

Therefore, the given two numbers are co- prime.

Question 6:

A number is divisible by both 5 and 12. By which other number will that number be

always divisible?

Answer:

Factors of 5 = 1, 5

Factors of 12 = 1, 2, 3, 4, 6, 12

As the common factor of these numbers is 1, the given two numbers are co- prime and

the number will also be divisible by their product, i.e. 60, and the factors of 60, i.e., 1,

2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60.

Question 7:

A number is divisible by 12. By what other number will that number be divisible?

Answer:

Since the number is divisible by 12, it will also be divisible by its factors i.e., 1, 2, 3, 4,

6, 12. Clearly, 1, 2, 3, 4, and 6 are numbers other than 12 by which this number is also

divisible.

Page 19 of 41


Class VI


Maths

Exercise 3.5

Question 1:

Which of the following statements are true?

(a) If a number is divisible by 3, it must be divisible by 9.

(b) If a number is divisible by 9, it must be divisible by 3.

(c) A number is divisible by 18, if it is divisible by both 3 and 6.

(d) If a number is divisible by 9 and 10 both, then it must be divisible by 90.

(e) If two numbers are co-primes, at least one of them must be prime.

(f) All numbers which are divisible by 4 must also be divisible by 8.

(g) All numbers which are divisible by 8 must also be divisible by 4.

(h) If a number exactly divides two numbers separately, it must exactly divide their

sum.

(i) If a number exactly divides the sum of two numbers, it must exactly divide the two

numbers separately.

Answer:

(a) False

6 is divisible by 3, but not by 9.

(b) True, as 9 = 3 × 3

Therefore, if a number is divisible by 9, then it will also be divisible by

3.

(c) False

30 is divisible by 3 and 6 both, but it is not divisible by 18.

(d) True, as 9 × 10 = 90

Therefore, if a number is divisible by 9 and 10 both, then it will also be divisible by 90.

(e) False

15 and 32 are co-primes and also composite.

(f) False

12 is divisible by 4, but not by 8.

(g) True, as 8 = 2 × 4

Therefore, if a number is divisible by 8, then it will also be divisible by 2 and 4.

(h) True

2 divides 4 and 8 as well as 12. (4 + 8 = 12)

(i) False

Page 20 of 41



Class VI


Maths

2 divides 12, but does not divide 7 and 5.

Question 2:

Here are two different factor trees for 60. Write the missing numbers.

(a)

(b)

Answer:

(a) As 6 = 2 × 3 and 10 = 5 × 2

(b) As 60 = 30 × 2, 30 = 10 × 3, and 10 = 5 × 2

Question 3:

Which factors are not included in the prime factorization of a composite number?

Answer:

1 and the number itself

Question 4:

Write the greatest 4-digit number and express it in terms of its prime factors.

Answer:

Greatest four-digit number = 9999

9999 = 3 × 3 × 11 × 101

Page 21 of 41



7 1729

13 247

19 19

1

Class VI


Maths

Question 5:

Write the smallest 5-digit number and express it in the form of its prime factors.

Answer:

Smallest five-digit number = 10,000

10000 = 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5

Question 6:

Find all prime factors of 1729 and arrange them in ascending order. Now state the

relation, if any; between two consecutive prime factors.

Answer:

1729 = 7 × 13 × 19

13 − 7 = 6, 19 − 13 = 6

Page 22 of 41


Class VI


Maths

The difference of two consecutive prime factors is 6.

Question 7:

The product of three consecutive numbers is always divisible by 6. Verify this statement

with the help of some examples.

Answer:

2 × 3 × 4 = 24, which is divisible by 6



Question 8:

The sum of two consecutive odd numbers is divisible by 4. Verify this statement with the

help of some examples.

Answer:

3 + 5 = 8, which is divisible by 4



Question 9:

In which of the following expressions, prime factorization has been done?

(a) 24 = 2 × 3 × 4 (b) 56 = 7 × 2 × 2 × 2

(c) 70 = 2 × 5 × 7 (d) 54 = 2 × 3 × 9

Answer:

(a) 24 = 2 × 3 × 4

Since 4 is composite, prime factorisation has not been done.

(b) 56 = 7 × 2 × 2 × 2

Since all the factors are prime, prime factorisation has been done.

(c) 70 = 2 × 5 × 7

Since all the factors are prime, prime factorisation has been done.

(d) 54 = 2 × 3 × 9

Since 9 is composite, prime factorisation has not been done.

Question 10:

Determine if 25110 is divisible by 45.

[Hint: 5 and 9 are co-prime numbers. Test the divisibility of the number by 5 and 9].

Answer:

45 = 5 × 9

Page 23 of 41



Class VI


Maths

Factors of 5 = 1, 5

Factors of 9 = 1, 3, 9

Therefore, 5 and 9 are co-prime numbers.

Since the last digit of 25110 is 0, it is divisible by 5.

Sum of the digits of 25110 = 2 + 5 + 1 + 1 + 0 = 9

As the sum of the digits of 25110 is divisible by 9, therefore, 25110 is divisible by 9.

Since the number is divisible by 5 and 9 both, it is divisible by 45.

Question 11:

18 is divisible by both 2 and 3. It is also divisible by 2 × 3 = 6. Similarly, a number is

divisible by both 4 and 6. Can we say that the number must also be divisible by 4 × 6 =

24? If not, give an example to justify our answer:

Answer:

No. It is not necessary because 12 and 36 are divisible by 4 and 6 both, but are not

divisible by 24.

Question 12:

I am the smallest number, having four different prime factors. Can you find me?

Answer:

Since it is the smallest number of such type, it will be the product of 4 smallest prime

numbers.

2 × 3 × 5 × 7 = 210

Page 24 of 41


2 18

3 9

3 3

1

2 48

2 24

2 12

2 6

3 3

1

Class VI


Maths

Exercise 3.6

Question 1:

Find the HCF of the following numbers:

(a) 18, 48 (b) 30, 42 (c) 18, 60

(d) 27, 63 (e) 36, 84 (f) 34, 102

(g) 70, 105, 175 (h) 91, 112, 49 (i) 18, 54, 81

(j) 12, 45, 75

Answer:

(a) 18, 48

18 = 2 × 3 × 3

48 = 2 × 2 × 2 × 2 × 3

HCF = 2 × 3 = 6

(b) 30, 42

2

30

Page 25 of 41



2 18

3 9

3 3

1

2 60

2 30

3 15

5 5

1

3 15

5 5

1

2 42

3 21

7 7

1

Class VI


Maths

30 = 2 × 3 × 5

42 = 2 × 3 × 7

HCF = 2 × 3 = 6

(c) 18, 60

18 = 2 × 3 × 3

60 = 2 × 2 × 3 × 5

Page 26 of 41


3 27

3 9

3 3

1

3 63

3 21

7 7

1

2 36

2 18

3 9

3 3

1

2 84

2 42

3 21

Class VI


Maths

HCF = 2 × 3 = 6

(d) 27, 63

27 = 3 × 3 × 3

63 = 3 × 3 × 7

HCF = 3 × 3 = 9

(e) 36, 84

Page 27 of 41


Class VI

36 = 2 × 2 × 3 × 3

84 = 2 × 2 × 3 × 7

HCF = 2 × 2 × 3 = 12

(f) 34, 102

2

34

17

17

1

2 102

3

51

17

17

1

34 = 2 × 17

102 = 2 × 3 × 17

HCF = 2 ×17 = 34

(g) 70, 105, 175

2

70

5

35

7

7

1

3 105


Page 28 of 41

Maths



7 7

1

7 91

13 13

1

2 112

2 56

2 28

2 14

7 7

1

5 35

7 7

1

5 175

5 35

7 7

1

Class VI


Maths

70 = 2 × 5 × 7

105 = 3 × 5 × 7

175 = 5 × 5 × 7

HCF = 5 × 7 = 35

(h) 91, 112, 49

Page 29 of 41


2 18

3 9

3 3

1

2 54

3 27

3 9

3 3

1

3 81

3 27

3 9

3 3

7 49

7 7

1

Class VI


Maths

91 = 7 × 13

112 = 2 × 2 × 2 × 2 × 7

49 = 7 × 7

HCF = 7

(i) 18, 54, 81

Page 30 of 41


2 12

2 6

3 3

1

3 45

3 15

5 5

1

3 75

5 25

5 5

1

Class VI


Maths

1

18 = 2 × 3 × 3

54 = 2 × 3 × 3 × 3

81 = 3 × 3 × 3 × 3

HCF = 3 × 3 = 9

(j) 12, 45, 75

12 = 2 ×2 × 3

45 = 3 × 3 × 5

75 = 3 × 5 × 5

HCF = 3

Question 2:

What is the HCF of two consecutive

Page 31 of 41


Class VI


Maths

(a) Numbers? (b) Even numbers? (c) Odd numbers?

Answer:

(i) 1 e.g., HCF of 2 and 3 is 1.

(ii) 2 e.g., HCF of 2 and 4 is 2.

(iii) 1 e.g., HCF of 3 and 5 is 1.

Question 3:

HCF of co-prime numbers 4 and 15 was found as follows by factorization:

4 = 2 × 2 and 15 = 3 × 5 since there is no common prime factors, so HCF of 4 and 15 is

0. Is the answer correct? If not, what is the correct HCF?

Answer:

No. The answer is not correct. 1 is the correct HCF.

Page 32 of 41


Class VI


Maths

Exercise 3.7

Question 1:

Renu purchases two bags of fertilizer of weight 75 kg and 69 kg. Find the maximum

value of weight which can measure the weight of the fertilizer exact number of times.

Answer:

Weight of the two bags = 75 kg and 69 kg

Maximum weight = HCF (75, 69)

3

5

5

3

23

75

25

5

1

69

23

1

75 = 3 × 5 × 5

69 = 3 × 23

HCF = 3

Hence, the maximum value of weight, which can measure the weight of the fertilizer

exact number of times, is 3 kg.

Question 2:

Three boys step off together from the same spot. Their steps measure 63 cm, 70 cm and

77 cm respectively. What is the minimum distance each should cover so that all can

cover the distance in complete steps?

Answer:

Step measure of 1st Boy = 63 cm

Step measure of 2nd Boy = 70 cm

Step measure of 3rd Boy = 77 cm

LCM of 63, 70, 77

Page 33 of 41


2 63, 70, 77

3 63, 35, 77

3 21, 35, 77

5 7, 35, 77

7 7, 7, 77

11 1, 1, 11

1, 1, 1

2 6, 8, 12

2 3, 4, 6

2 3, 2, 3

Class VI


Maths

LCM = 2 × 3 × 3 × 5 × 7 × 11 = 6930

Hence, the minimum distance each should cover so that all can cover the distance in

complete steps is 6930 cm.

Question 3:

The length, breadth and height of a room are 825 cm, 675 cm and 450 cm respectively.

Find the longest tape which can measure the three dimensions of the room exactly.

Answer:

Length = 825 cm = 3 × 5 × 5 × 11

Breadth = 675 cm = 3 × 3 × 3 × 5 × 5

Height = 450 cm = 2 × 3 × 3 × 5 × 5

Longest tape = HCF of 825, 675, and 450 = 3 × 5 × 5 = 75 cm

Therefore, the longest tape is 75 cm.

Question 4:

Determine the smallest 3-digit number which is exactly divisible by 6, 8 and 12.

Answer:

Smallest number = LCM of 6, 8, 12

Page 34 of 41


2 8, 10, 12

2 4, 5, 6

2 2, 5, 3

3 1, 5, 3

5 1, 5, 1

1,1,1

3 3, 1, 3

1, 1, 1

Class VI


Maths

LCM = 2 × 2 × 2 × 3 = 24

We have to find the smallest 3-digit multiple of 24.

It can be seen that 24 × 4 = 96 and 24 × 5 = 120.

Hence, the smallest 3-digit number which is exactly divisible by 6, 8, and 12 is 120.

Question 5:

Determine the greatest 3-digit number exactly divisible by 8, 10 and 12.

Answer:

LCM of 8, 10, and 12

LCM = 2 × 2 × 2 × 3 × 5 = 120

We have to find the greatest 3-digit multiple of 120.

It can be seen that 120 ×8 = 960 and 120 × 9 = 1080.

Hence, the greatest 3-digit number exactly divisible by 8, 10, and 12 is 960.

Question 6:

The traffic lights at three different road crossings change after every 48 seconds, 72

seconds and 108 seconds respectively. If they change simultaneously at 7 a.m., at what

time will they change simultaneously again?

Answer:

Time period after which these lights will change = LCM of 48, 72, 108

Page 35 of 41


2 48, 72, 108

2 24, 36, 54

2 12, 18, 27

2 6, 9, 27

3 3, 9, 27

3 1, 3, 9

3 1, 1, 3

1, 1, 1

Class VI


Maths

LCM = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 432

They will change together after every 432 seconds i.e., 7 min 12 seconds.

Hence, they will change simultaneously at 7:07:12 am.

Question 7:

Three tankers contain 403 litres, 434 litres and 465 litres of diesel respectively. Find the

maximum capacity of a container that can measure the diesel of the three containers

exact number of times.

Answer:

Maximum capacity of the required tanker = HCF of 403, 434, 465

403 = 13 × 31

434 = 2 × 7 × 31

465 = 3 × 5 × 31

HCF = 31

∴A container of capacity 31 l can measure the diesel of 3 containers exact number of

times

Page 36 of 41


2 18, 24, 32

2 9, 12, 16

2 9, 6, 8

2 9, 3, 4

2 9, 3, 2

3 9, 3, 1

3 3, 1, 1

1, 1, 1

2 6, 15, 18

3 3, 15, 9

3 1, 5, 3

5 1, 5, 1

1, 1, 1

Class VI


Maths

Question 8:

Find the least number which when divided by 6, 15 and 18 leave remainder 5 in each

case.

Answer:

LCM of 6, 15, 18

LCM = 2 × 3 × 3 × 5 = 90

Required number = 90 + 5 = 95

Question 9:

Find the smallest 4-digit number which is divisible by 18, 24 and 32.

Answer:

LCM of 18, 24, and 32

Page 37 of 41


2 12, 5

2 6, 5

3 3, 5

5 1, 5

1, 1

2 9, 4

2 9, 2

3 9, 1

3 3, 1

1, 1

Class VI


Maths

LCM = 2 × 2 × 2 × 2 × 2 × 3 × 3 = 288

We have to find the smallest 4-digit multiple of 288.

It can be observed that 288 ×3 = 864 and 288 ×4 = 1152.

Therefore, the smallest 4-digit number which is divisible by 18, 24, and 32 is

1152.

Question 10:

Find the LCM of the following numbers:

(a) 9 and 4 (b) 12 and 5

(c) 6 and 5 (d) 15 and 4

Observe a common property in the obtained LCMs. Is LCM the product of two numbers in

each case?

Answer:

(a)

LCM = 2 × 2 × 3 × 3 = 36

(b)

LCM = 2 × 2 × 3 × 5 = 60

Page 38 of 41


2 15, 4

2 15, 2

3 15, 1

5 5, 1

1, 1

2 6, 5

3 3, 5

5 1, 5

1, 1

2 5, 20

2 5, 10

Class VI


Maths

(c)

LCM = 2 × 3 × 5 = 30

(d)

LCM = 2 × 2 × 3 × 5 = 60

Yes, it can be observed that in each case, the LCM of the given numbers is the product

of these numbers. When two numbers are co-prime, their LCM is the product of those

numbers. Also, in each case, LCM is a multiple of 3.

Question 11:

Find the LCM of the following numbers in which one number is the factor of the other.

(a) 5, 20 (b) 6, 18

(c) 12, 48 (d) 9, 45

What do you observe in the results obtained?

Answer:

(a)

Page 39 of 41


2 12, 48

2 6, 24

2 3, 12

2 3, 6

3 3, 3

1, 1

3 9, 45

3 3, 15

5 1, 5

1, 1

2 6, 18

3 3, 9

3 1, 3

1, 1

5 5, 5

1, 1

Class VI


Maths

LCM = 2 × 2 × 5 = 20

(b)

LCM = 2 × 3 × 3 = 18

(c)

LCM = 2 × 2 × 2 × 2 × 3 = 48

(d)

Page 40 of 41


Class VI


Maths

LCM = 3 × 3 × 5 = 45

Yes, it can be observed that in each case, the LCM of the given numbers is the larger

number. When one number is a factor of the other number, their LCM will be the larger

number.

Page 41 of 41


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